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# The operation x#n for all positive integers greater than 1

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Senior Manager
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The operation x#n for all positive integers greater than 1 [#permalink]

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13 Aug 2010, 09:51
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Difficulty:

95% (hard)

Question Stats:

42% (06:14) correct 58% (02:30) wrong based on 428 sessions

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The operation x#n for all positive integers greater than 1 is defined in the following manner: x#n = x to the power of x#(n-1)

If x#1 = x, which of the following expressions has the greatest value?

A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3
[Reveal] Spoiler: OA

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Re: Just 800 Level question [#permalink]

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13 Aug 2010, 10:33
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Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?

A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3

Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.

Back to the original question:
Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: $$x@n=x^{(x@(n-1))}$$ and $$x@1=x$$:
______________________________________

$$x@2=x^{(x@1)}=x^x$$, as $$x@1=x$$;
$$x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}$$;
$$x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}$$;
...
Basically n in x@n represents the # of stacked x-es.

A. $$(3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}$$

B. $$3@(1@3)=3@(1^{1^1})=3@1=3$$

C. $$(2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}$$

D. $$2@(2@3)=2@(2^{2^2})=2@16$$ this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=2^2^2^2^2^2^2^2^2^2^2^2^2^16=.... ;

E. $$(2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}$$

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

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Re: Just 800 Level question [#permalink]

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15 Aug 2010, 21:39
Where'd this question come from?

And as always -- well done Bunuel!

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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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21 Jun 2013, 02:46
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61

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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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21 Jun 2013, 21:29
1
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Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61

Well originally, i got this question wrong.
Bu then i figured an alternative approach which would have been very easy to start with, please let me know if there is something wrong with it.

First of all to look at the options, everything looks to be in the form of either,

a@(b@c) or (a@b)@c

Now lets quickly try to see the diff between the magnitude of these different forms, lets take a,b,c = 2 for comparison purpose.

2@(2@2) = 2@4 = 2^(2^(2^2)) = 2^16
while, (2@2)@2 = (2^2)@2 = 4@2 = 4^4 = 2^8

So well, there is a huge magnitude diff between the post, and option 1 def has much more wightage.

now, lets late at our options in the question : (all the options have either 2 or 3 as the values for a, b and c in a@(b@c) and (a@b)@c, so we can make a direct comparison.

We have B and D in a@(b@c) form, however since B has 1 as one of the numbers ( 1 and 0 are special no's with special properties),

so we can easily see that 3@(1@3) = 3@1 = 3 (which is most definitely the smallest)

Which leaves to D (2@16) now which is the answer.
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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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23 Jun 2013, 05:51
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61

Hi Bunuel,

Could you please let me know if the above approach is fine? or if there is something wrong with it?
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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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24 Jun 2013, 01:12
Sir, I am totally confused with wat I have to study and on wat questions I should practice in this website. There are so many navigation from 1 page to other. Please provide me the solution where I can improve myself with enough practice.

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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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24 Jun 2013, 01:35
Bhagavantha wrote:
Sir, I am totally confused with wat I have to study and on wat questions I should practice in this website. There are so many navigation from 1 page to other. Please provide me the solution where I can improve myself with enough practice.

Hope it helps.
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Re: Just 800 Level question [#permalink]

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22 Oct 2013, 22:46
Bunuel wrote:
Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?

A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3

Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.

Back to the original question:
Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: $$x@n=x^{(x@(n-1))}$$ and $$x@1=x$$:
______________________________________

$$x@2=x^{(x@1)}=x^x$$, as $$x@1=x$$;
$$x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}$$;
$$x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}$$;
...
Basically n in x@n represents the # of stacked x-es.

A. $$(3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}$$

B. $$3@(1@3)=3@(1^{1^1})=3@1=3$$

C. $$(2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}$$

D. $$2@(2@3)=2@(2^{2^2})=2@16$$ this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=2^2^2^2^2^2^2^2^2^2^2^2^2^16=.... ;

E. $$(2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}$$

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation?

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Re: Just 800 Level question [#permalink]

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22 Oct 2013, 23:16
kartboybo wrote:
Bunuel wrote:
Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?

A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3

Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.

Back to the original question:
Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: $$x@n=x^{(x@(n-1))}$$ and $$x@1=x$$:
______________________________________

$$x@2=x^{(x@1)}=x^x$$, as $$x@1=x$$;
$$x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}$$;
$$x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}$$;
...
Basically n in x@n represents the # of stacked x-es.

A. $$(3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}$$

B. $$3@(1@3)=3@(1^{1^1})=3@1=3$$

C. $$(2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}$$

D. $$2@(2@3)=2@(2^{2^2})=2@16$$ this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=2^2^2^2^2^2^2^2^2^2^2^2^2^16=.... ;

E. $$(2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}$$

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation?

Don't know what more can I add:
We have that $$x@2=x^x$$, thus $$x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}$$.
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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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31 Oct 2013, 08:48
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It took me a while to figure out, but then I am glad in the end.

I just have a diff approach compared to insanely awesome Bunuel's, so thought to share.

We are given x#1 = 1 . Given x > 1, so # could be either of ^ , / , *

So if we start solving x#1, as per the original expression i.e x#n = .....

We arrive to x#0 = 1 in the second setup. What could be # then ?
The only option is # is exponential.

Now its school stuff to know which one is greater, Option D clearly stands out.

A. (3^2)^2 = 81
B. 3
C. 2^6
D. 2^8
E. 2^6.

Having given GMAT once, I think this is a 700 question, easy, but looks intimidating in beginning

Liked it ? Appreciate it

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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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01 Nov 2014, 18:04
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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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12 Nov 2015, 22:42
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Re: The operation x#n for all positive integers greater than 1 [#permalink]

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10 Jan 2017, 13:58
Hello from the GMAT Club BumpBot!

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Re: The operation x#n for all positive integers greater than 1   [#permalink] 10 Jan 2017, 13:58
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